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Let $f \in C^1([0;1],\mathbb{R})$ such that $f(0)=0$. $$\text{Prove that} \qquad \int_{0}^{1}\left ( \frac{f(x)}{x} \right )^2dx \leq 4\int_{0}^{1}(f'(x))^2dx$$

My attempt:
Let $$g(x)=\begin{cases} \frac{f(x)}{x^\alpha}, & \quad x\neq 0\\ \frac{f^{(\alpha)}(0)}{\alpha!}, & \quad x=0 \end{cases},$$ where $\alpha \in \mathbb{N}$ and $g \in C^1([0;1],\mathbb{R})$. We have $$ \begin{split} f(x)&=x^\alpha g(x)\\ f'(x)&=\alpha x^{\alpha-1}g(x)+x^\alpha g'(x)\\ I &= 4\int_{0}^{1}(f'(x))^2dx \\ &= 4\left ( \int_{0}^{1}\alpha^2x^{2\alpha-2}(g(x))^2dx + \int_{0}^{1}x^{2\alpha}(g'(x))^2dx + \int_{0}^{1}2g(x)g'(x)\alpha x^{2\alpha-1}dx \right )\\ &= 4\left ( \int_{0}^{1}\alpha^2x^{2\alpha-2}(g(x))^2dx + \int_{0}^{1}x^{2\alpha}(g'(x))^2dx + \int_{0}^{1}((g(x))^2)'\alpha x^{2\alpha-1}dx \right )\\ &= 4\left ( \int_{0}^{1}\alpha^2x^{2\alpha-2}(g(x))^2dx + \int_{0}^{1}x^{2\alpha}(g'(x))^2dx + \alpha (g(1))^2\\ - \int_{0}^{1}(g(x))^2\alpha(2\alpha-1)x^{2\alpha-2}dx \right )\\ & \text{ (integrate by parts)}\\ &= 4\left ( \int_{0}^{1}\alpha(1-\alpha)x^{2\alpha-2}(g(x))^2dx + \int_{0}^{1}x^{2\alpha}(g'(x))^2dx + \alpha (g(1))^2 \right )\\ &= 4\left ( \alpha(1-\alpha)\int_{0}^{1}\left ( \frac{f(x)}{x} \right )^2dx + \int_{0}^{1}x^{2\alpha}(g'(x))^2dx + \alpha (g(1))^2 \right )\\ &\geq 4\alpha(1-\alpha)\int_{0}^{1}\left ( \frac{f(x)}{x} \right )^2dx \end{split} $$ I'm stuck here because $\alpha(1-\alpha) \leq 0$ for $\alpha \in \mathbb{N}$. Do you have any idea? Thank you in advance.

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  • $\begingroup$ What's $α$ for?? $\endgroup$ – Saad Apr 15 at 13:43
  • $\begingroup$ If $f(x)=xg(x)$ then $$4\int_{0}^{1}(f'(x))^2dx = 4\left ( \int_{0}^{1}x^2(g'(x))^2dx + (g(1))^2\right )$$ $\endgroup$ – Tinh Tran Apr 15 at 13:48
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I would suggest this instead: assume $f$ is not identcally zero, and let $0 < \epsilon < 1$ be sufficiently small that $\displaystyle \int_\epsilon^1 \frac 1{x^2} f(x)^2 \, dx > 0.$ Integrate by parts to make the estimate $$ \int_{\epsilon}^1 \frac 1{x^2} f(x)^2 \, dx = \frac{f(\epsilon)^2}{\epsilon} - \frac{f(1)^2}{1} + \int_{\epsilon}^1 \frac 1x 2 f(x) f'(x) \, dx \le \frac{f(\epsilon)^2}{\epsilon} + \int_{\epsilon}^1 \frac 1x 2 f(x) f'(x) \, dx.$$

The Cauchy-Schwarz inequality gives you $$ \int_{\epsilon}^1 \frac 1x 2 f(x) f'(x) \, dx \le 2 \left( \int_{\epsilon}^1 \frac 1{x^2} f(x)^2 \, dx \right)^{1/2} \left( \int_{\epsilon}^1 f'(x)^2 \, dx \right)^{1/2}$$ so you can divide to obtain $$\left( \int_{\epsilon}^1 \frac 1{x^2} f(x)^2 \, dx \right)^{1/2} \le \left( \int_{\epsilon}^1 \frac 1{x^2} f(x)^2 \, dx \right)^{-1/2} \frac{f(\epsilon)^2}{\epsilon} + 2\left( \int_{\epsilon}^1 f'(x)^2 \, dx \right)^{1/2}.$$

There are two things to note: the expression $\displaystyle \left( \int_{\epsilon}^1 \frac 1{x^2} f(x)^2 \, dx \right)^{-1/2}$ is decreasing as $\epsilon \to 0^+$, and $\displaystyle \frac{f(\epsilon)^2}{\epsilon} \le \int_0^\epsilon f'(t)^2 \, dt \to 0$ as $\epsilon \to 0^+$. Thus $$ \left( \int_{\epsilon}^1 \frac 1{x^2} f(x)^2 \, dx \right)^{-1/2} \frac{f(\epsilon)^2}{\epsilon} \to 0$$ as $\epsilon \to 0^+$. Take the limit in all three terms above to find $$ \left( \int_0^1 \frac 1{x^2} f(x)^2 \, dx \right)^{1/2} \le 2 \left( \int_0^1 f'(x)^2 \, dx \right)^{1/2}.$$

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